What are the solutions to the equation (x-8)^2+x^2=(x+8)^2 ?

The solutions to the equation (x-8)^2+x^2=(x+8)^2 are x=32,x=0

Find the zeros of (x-8)^2+x^2=(x+8)^2

The zeros of (x-8)^2+x^2=(x+8)^2 are x=32,x=0

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(x - 8)^{2} + x^{2} = (x + 8)^{2}

x = 32, x = 0

Solution steps

(x - 8)^{2} + x^{2} = (x + 8)^{2}

Expand

(x - 8)^{2} + x^{2} : 2 x^{2} - 16 x + 64

Expand

(x + 8)^{2} : x^{2} + 16 x + 64

2 x^{2} - 16 x + 64 = x^{2} + 16 x + 64

Move

64

to the left side

2 x^{2} - 16 x = x^{2} + 16 x

Move

16 x

to the left side

2 x^{2} - 32 x = x^{2}

Move

x^{2}

to the left side

x^{2} - 32 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

(- 32)^{2} - 4 \cdot 1 \cdot 0 = 32

x_{1, 2} = \frac{- ( - 32 ) \pm 32}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 32 ) + 32}{2 \cdot 1}, x_{2} = \frac{- ( - 32 ) - 32}{2 \cdot 1}

x = \frac{- ( - 32 ) + 32}{2 \cdot 1} : 32

x = \frac{- ( - 32 ) - 32}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

x = 32, x = 0

\frac{a + 6}{3} = \frac{2 a + 4}{5}

\frac{2}{5} x - 6 = 8

\frac{x}{3} + \frac{2}{3} = \frac{5}{3}

x^{2} + 10 x + 61 = 0

- 8 \leq 3 x - 5 \leq 7

What are the solutions to the equation (x-8)^2+x^2=(x+8)^2 ?
The solutions to the equation (x-8)^2+x^2=(x+8)^2 are x=32,x=0
Find the zeros of (x-8)^2+x^2=(x+8)^2
The zeros of (x-8)^2+x^2=(x+8)^2 are x=32,x=0